# To what volume should 5.0 g of KCl be diluted in order to prepare 0.25 M solution?

Mar 5, 2016

0.27 L

#### Explanation:

We are looking for volume and are given a mass and a molarity.

Note:

$M = \frac{m o {l}_{s o l u t e}}{L i t e r {s}_{s o l u t i o n}}$

Rearranging this equation to solve for volume:

${L}_{s o l u t i o n} = \frac{m o {l}_{s o l u t e}}{M}$

We have M, but we need $m o {l}_{s o l u t e}$. To get this, we use the molecular weight of KCl, which is:

$74.55 {g}_{K C l} / \left(m o {l}_{K C l}\right)$

To get the moles of KCl:

$5.0 {g}_{K C l} \cdot \frac{1 m o {l}_{K C l}}{74.55 {g}_{K C l}} = 0.067 m o {l}_{K C l}$

Finally, plug back into the equation for $L i t e r {s}_{s o l u t i o n}$:
${L}_{s o l u t i o n} = \frac{m o {l}_{s o l u t e}}{M}$
L_(solution)=(0.067mol_(KCl))/(0.25*(mol_(KCl))/(L_(solution))
${L}_{s o l u t i o n} = 0.27 L$

Note: This is the total solution volume, not the volume of solvent. Total volume of solute + solvent is the solution volume.